Largest product of a subarray of size k

Given an array consisting of n positive integers, and an integer k. Find the largest product subarray of size k, i.e., find maximum produce of k contiguous elements in the array where k <= n.

Examples :

Input: arr[] = {1, 5, 9, 8, 2, 4,
                 1, 8, 1, 2} 
       k = 6
Output:   4608  
The subarray is {9, 8, 2, 4, 1, 8}

Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}
       k = 4
Output:   720  
The subarray is {5, 9, 8, 2}

Input: arr[] = {2, 5, 8, 1, 1, 3};
       k = 3             
Output:   80  
The subarray is {2, 5, 8}

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Simple : O(n*k))
A Naive approach would be to consider all the subarrays of size k one by one. Such a approach would require two loops hence the complexity would be O(n*k).

Method 2 (Efficient : O(n))
We can solve it in O(n) by using the fact that product of a subarray of size k can be computed in O(1) time if we have product of previous subarray available with us.

curr_product = (prev_product / arr[i-1]) * arr[i + k -1]

prev_product : Product of subarray of size k beginning 
               with arr[i-1]

curr_product : Product of subarray of size k beginning 
               with arr[i]

In this way we can compute the maximum k size subarray product in only one traversal. Below is C++ implementation of the idea.

C++

// C++ program to find the maximum product of a subarray 
// of size k. 
#include <bits/stdc++.h> 
using namespace std; 
  
// This function returns maximum product of a subarray 
// of size k in given arrar, arr[0..n-1]. This function 
// assumes that k is smaller than or equal to n. 
int findMaxProduct(int arr[], int n, int k) 
{ 
    // Initialize the MaxProduct to 1, as all elements 
    // in the array are positive 
    int MaxProduct = 1; 
    for (int i=0; i<k; i++) 
        MaxProduct *= arr[i]; 
  
    int prev_product = MaxProduct; 
  
    // Consider every product beginning with arr[i] 
    // where i varies from 1 to n-k-1 
    for (int i=1; i<=n-k; i++) 
    { 
        int curr_product = (prev_product/arr[i-1]) * 
                            arr[i+k-1]; 
        MaxProduct = max(MaxProduct, curr_product); 
        prev_product = curr_product; 
    } 
  
    // Return the maximum product found 
    return MaxProduct; 
} 
  
// Driver code 
int main() 
{ 
    int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}; 
    int k = 6; 
    int n = sizeof(arr1)/sizeof(arr1[0]); 
    cout << findMaxProduct(arr1, n, k) << endl; 
  
    k = 4; 
    cout << findMaxProduct(arr1, n, k) << endl; 
  
    int arr2[] = {2, 5, 8, 1, 1, 3}; 
    k = 3; 
    n = sizeof(arr2)/sizeof(arr2[0]); 
    cout << findMaxProduct(arr2, n, k); 
  
    return 0; 
} 

Java

// Java program to find the maximum product of a subarray 
// of size k 
import java.io.*; 
import java.util.*; 
  
class GFG  
{ 
    // Function returns maximum product of a subarray 
    // of size k in given arrar, arr[0..n-1]. This function 
    // assumes that k is smaller than or equal to n. 
    static int findMaxProduct(int arr[], int n, int k) 
    { 
        // Initialize the MaxProduct to 1, as all elements 
        // in the array are positive 
        int MaxProduct = 1; 
        for (int i=0; i<k; i++) 
            MaxProduct *= arr[i]; 
   
        int prev_product = MaxProduct; 
   
        // Consider every product beginning with arr[i] 
        // where i varies from 1 to n-k-1 
        for (int i=1; i<=n-k; i++) 
        { 
            int curr_product = (prev_product/arr[i-1]) * 
                                arr[i+k-1]; 
            MaxProduct = Math.max(MaxProduct, curr_product); 
            prev_product = curr_product; 
        } 
   
        // Return the maximum product found 
        return MaxProduct; 
    } 
      
    // driver program 
    public static void main (String[] args)  
    { 
        int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}; 
        int k = 6; 
        int n = arr1.length; 
        System.out.println(findMaxProduct(arr1, n, k)); 
   
        k = 4; 
        System.out.println(findMaxProduct(arr1, n, k)); 
   
        int arr2[] = {2, 5, 8, 1, 1, 3}; 
        k = 3; 
        n = arr2.length; 
        System.out.println(findMaxProduct(arr2, n, k)); 
    } 
} 
  
// This code is contributed by Pramod Kumar 

Python3

# Python 3 program to find the maximum  
# product of a subarray of size k. 
  
# This function returns maximum product  
# of a subarray of size k in given arrar, 
# arr[0..n-1]. This function assumes  
# that k is smaller than or equal to n. 
def findMaxProduct(arr, n, k) : 
    
    # Initialize the MaxProduct to 1,  
    # as all elements in the array  
    # are positive 
    MaxProduct = 1
    for i in range(0, k) : 
        MaxProduct = MaxProduct * arr[i] 
          
    prev_product = MaxProduct 
   
    # Consider every product beginning 
    # with arr[i] where i varies from 
    # 1 to n-k-1 
    for i in range(1, n - k + 1) : 
        curr_product = (prev_product // arr[i-1]) * arr[i+k-1] 
        MaxProduct = max(MaxProduct, curr_product) 
        prev_product = curr_product 
      
      
    # Return the maximum product found 
    return MaxProduct 
      
# Driver code 
arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2] 
k = 6
n = len(arr1) 
print (findMaxProduct(arr1, n, k) ) 
k = 4
print (findMaxProduct(arr1, n, k)) 
  
arr2 = [2, 5, 8, 1, 1, 3] 
k = 3
n = len(arr2) 
  
print(findMaxProduct(arr2, n, k)) 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to find the maximum  
// product of a subarray of size k 
using System; 
  
class GFG  
{ 
    // Function returns maximum  
    // product of a subarray of  
    // size k in given arrar,  
    // arr[0..n-1]. This function  
    // assumes that k is smaller  
    // than or equal to n. 
    static int findMaxProduct(int []arr,  
                              int n, int k) 
    { 
        // Initialize the MaxProduct  
        // to 1, as all elements 
        // in the array are positive 
        int MaxProduct = 1; 
        for (int i = 0; i < k; i++) 
            MaxProduct *= arr[i]; 
  
        int prev_product = MaxProduct; 
  
        // Consider every product beginning  
        // with arr[i] where i varies from  
        // 1 to n-k-1 
        for (int i = 1; i <= n - k; i++) 
        { 
            int curr_product = (prev_product /  
                                 arr[i - 1]) *  
                                 arr[i + k - 1]; 
            MaxProduct = Math.Max(MaxProduct,  
                                  curr_product); 
            prev_product = curr_product; 
        } 
  
        // Return the maximum 
        // product found 
        return MaxProduct; 
    } 
      
    // Driver Code 
    public static void Main ()  
    { 
        int []arr1 = {1, 5, 9, 8, 2,  
                      4, 1, 8, 1, 2}; 
        int k = 6; 
        int n = arr1.Length; 
        Console.WriteLine(findMaxProduct(arr1, n, k)); 
  
        k = 4; 
        Console.WriteLine(findMaxProduct(arr1, n, k)); 
  
        int []arr2 = {2, 5, 8, 1, 1, 3}; 
        k = 3; 
        n = arr2.Length; 
        Console.WriteLine(findMaxProduct(arr2, n, k)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find the maximum  
// product of a subarray of size k. 
  
// This function returns maximum  
// product of a subarray of size  
// k in given arrar, arr[0..n-1]. 
// This function assumes that k  
// is smaller than or equal to n. 
function findMaxProduct( $arr, $n, $k) 
{ 
      
    // Initialize the MaxProduct to 
    // 1, as all elements 
    // in the array are positive 
    $MaxProduct = 1; 
    for($i = 0; $i < $k; $i++) 
        $MaxProduct *= $arr[$i]; 
  
    $prev_product = $MaxProduct; 
  
    // Consider every product 
    // beginning with arr[i] 
    // where i varies from 1  
    // to n-k-1 
    for($i = 1; $i < $n - $k; $i++) 
    { 
        $curr_product = ($prev_product / $arr[$i - 1]) * 
                                       $arr[$i + $k - 1]; 
        $MaxProduct = max($MaxProduct, $curr_product); 
        $prev_product = $curr_product; 
    } 
  
    // Return the maximum 
    // product found 
    return $MaxProduct; 
} 
  
    // Driver code 
    $arr1 = array(1, 5, 9, 8, 2, 4, 1, 8, 1, 2); 
    $k = 6; 
    $n = count($arr1); 
    echo findMaxProduct($arr1, $n, $k),"\n" ; 
  
    $k = 4; 
    echo findMaxProduct($arr1, $n, $k),"\n"; 
  
    $arr2 = array(2, 5, 8, 1, 1, 3); 
    $k = 3; 
    $n = count($arr2); 
    echo findMaxProduct($arr2, $n, $k); 
  
// This code is contributed by anuj_67. 
?> 

Output :

4608
720
80

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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